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Write $r=|V_0|$, $s=|V_1|$, and $e=|E(G)|=\alpha |X||Y|$. We construct a random graph homomorphism $\Psi :T\to G$. Choose a root of the tree. Map it according to the degree-biased vertex distribution on the corresponding side of the bipartite graph; after mapping any vertex, map each child independently and uniformly to a neighbour of its parent's image.

Every oriented tree edge is then mapped uniformly onto the $e$ edges of $G$. Every tree vertex in $V_0$ has the degree-biased marginal on $X$, of entropy $H_X$, and every vertex in $V_1$ has the analogous marginal of entropy $H_Y$. Repeated use of the chain rule for information entropy along the rooted tree givesSince a $k$-vertex tree has $k-1$ edges,The maximum entropy distribution on a finite set gives $H_X\le \log |X|$ and $H_Y\le \log |Y|$. HenceIf $N$ is the number of bipartition-respecting graph homomorphisms $T\to G$, the support of $\Psi$ has size $N$, so the maximum entropy distribution on a finite set also gives $H(\Psi )\le \log N$. ThusThere are $|X{|}^r|Y{|}^s$ bipartition-respecting maps in total, so a uniformly chosen one is a graph homomorphism with probability at least ${\alpha }^{k-1}$. This proves the Sidorenko inequality for trees.

Solved by gpt-5.6-sol high.

Letwhere $\phi$ is the golden ratio. If the union-closed family consists of one nonempty set, any element of that set has frequency one, so assume its cardinality exceeds one. Choose independent uniform members $A,B$ and let $X,Y\in \{0,1{\}}^n$ be their characteristic vectors of sets. Then $H(X)=H(Y)=\log |\mathcal{A}|>0$.

Suppose for a contradiction that every element has frequency $p_i<c$. Put $q_i=1-p_i>1/\phi$, and let $Z=X\mathrm{OR}Y$, the characteristic vector of a set of $A\cup B$. The chain rule for information entropy and conditioning reduces entropy givebecause $Z_{<i}$ is a function of $(X_{<i},Y_{<i})$.

Fix the two prefixes and setThe two conditioned bits are independent random variables, and $Z_i=0$ exactly when both are zero. The supplied binary entropy product inequality therefore givesAveraging over the independent prefixes yieldswhenever either conditional entropy is positive. Summing and using $H(X)>0$ gives $H(Z)>H(X)$.

But set union keeps $A\cup B$ inside the union-closed family, so $Z$ is supported on $\mathcal{A}$. The maximum entropy distribution on a finite set gives $H(Z)\le \log |\mathcal{A}|=H(X)$, a contradiction. Some element must therefore occur in at least $c|\mathcal{A}|$ members, proving the entropy bound for a union-closed family.

Solved by gpt-5.6-sol high.

Monotonicity of the ordered probabilities gives $i{p}_i\le {\sum }_{j\le i}{p}_j\le 1$, so $\log i\le -\log p_i$. ConsequentlyGiven $L^{\ast }(X)=l$, the source symbol can take at most $2^l$ values. The maximum entropy distribution on a finite set is uniform, henceAveraging this conditional entropy inequality proves

Solved by gpt-5.6-sol high.